Return to Answer

For a particular tuple of values, we test it by constructing a directed graph of all tuples smaller then it, element-wise, representing the distance from the last element of each series, with the edges being "increase all values by 1, except for one which is set to 0", and then testing whether there's a cycle in the graph.

For example, for $(2, 4, 4)$, we get this graph:

Note the $(1,0,2), (0, 1, 3), (1, 2,0), (0, 3, 1)$ cycle, which gives the solution $1312$ repeating (each number is the index of the series the value occurs in)

This immediately proves that $\sum_i {\frac1{N_i}} \leq 1$ is necessary, otherwise $f(p) = \sum_i {\frac{p_i}{N_i}}$ gives a topological sort of the graph.

I have programmed this, and for $n \leq 5$, a tuple $N_1, N_2, ..., N_n$ is valid if it's elementwise greater than or equal to any of the following:

$n=1$: $$(2)$$

$n=2$: $$(2, 2)$$

$n=3$: $$(2, 4, 4), (3, 3, 3)$$

$n=4$: $$(2, 4, 8, 8), (2,6,6,6),(3,3,6,6),(3,4,5,8),(3,5,5,5),(4,4,4,4)$$

$n=5$: $$(2, 4, 8, 16, 16), (2, 4, 12, 12, 12), (2, 6, 6, 12, 12), (2, 6, 8, 10, 16), (2, 6, 10, 10, 10), (2, 8, 8, 8, 8), (3, 3, 6, 12, 12), (3, 3, 9, 9, 9), (3, 4, 5, 14, 14), (3, 4, 6, 10, 16), (3, 4, 6, 11, 11), (3, 4, 8, 8, 8), (3, 5, 5, 9, 9), (3, 5, 6, 7, 12), (3, 5, 7, 7, 9), (3, 5, 7, 8, 8), (3, 6, 6, 6, 6), (4, 4, 4, 8, 8), (4, 4, 5, 7, 12), (4, 4, 6,6, 6), (4, 5, 5, 6, 10), (4, 5, 5, 7, 7), (5, 5, 5, 5, 5)$$

$n=6$: $$(2, 4, 8, 16, 32, 32), (2, 4, 8, 24, 24, 24), (2, 4, 12, 12, 24, 24), (2, 4, 12, 20, 20, 20), (2, 4, 16, 16, 16, 16), (2, 6, 6, 12, 24, 24), (2, 6, 6, 18, 18, 18), (2, 6, 8, 10, 28, 28), (2, 6, 8, 12, 20, 32), (2, 6, 8, 12, 22, 22), (2, 6, 8, 16, 16, 16), (2, 6, 12, 12, 12, 12), (2, 10, 10, 10, 10, 10), (3, 3, 6, 12, 24, 24), (3, 3, 6, 18, 18, 18), (3, 3, 9, 9, 18, 18), (3, 3, 9, 15, 15, 15), (3, 3, 12, 12, 12, 12), (3, 4, 5, 14, 28, 28), (3, 4, 5, 22, 22, 22), (3, 4, 6, 10, 32, 32), (3, 4, 6, 11, 21, 32), (3, 4, 6, 11, 22, 22), (3, 4, 6, 16, 16, 16), (3, 4, 8, 8, 16, 16), (3, 4, 8, 13, 13, 13), (3, 4, 11, 11, 11, 11), (3, 5, 5, 9, 18, 18), (3, 5, 5, 14, 14, 14), (3, 5, 6, 7, 24, 24), (3, 5, 6, 8, 21, 21), (3, 5, 6, 9, 15, 24), (3, 5, 6, 9, 17, 17), (3, 5, 6, 12, 12, 12), (3, 5, 9, 9, 9, 9), (3, 8, 8, 8, 8, 8), (4, 4, 4, 8, 16, 16), (4, 4, 4, 12, 12, 12), (4, 4, 6, 6, 12, 12), (4, 4, 6, 10, 10, 10), (4, 4, 8, 8, 8, 8), (4, 7, 7, 7, 7, 7), (6, 6, 6, 6, 6, 6)$$

I can't see any clear pattern which would allow a simpler solution, but it's definitely possible one exists.

For a particular tuple of values, we test it by constructing a directed graph of all tuples smaller then it, element-wise, representing the distance from the last element of each series, with the edges being "increase all values by 1, except for one which is set to 0", and then testing whether there's a cycle in the graph.

For example, for $(2, 4, 4)$, we get this graph:

Note the $(1,0,2), (0, 1, 3), (1, 2,0), (0, 3, 1)$ cycle, which gives the solution $1312$ repeating (each number is the index of the series the value occurs in)

This immediately proves that $\sum_i {\frac1{N_i}} \leq 1$ is necessary, otherwise $f(p) = \sum_i {\frac{p_i}{N_i}}$ gives a topological sort of the graph.

I have programmed this, and for $n \leq 6$, a tuple $N_1, N_2, ..., N_n$ is valid if it's elementwise greater than or equal to any of the following:

$n=1$: $$(2)$$

$n=2$: $$(2, 2)$$

$n=3$: $$(2, 4, 4), (3, 3, 3)$$

$n=4$: $$(2, 4, 8, 8), (2,6,6,6),(3,3,6,6),(3,4,5,8),(3,5,5,5),(4,4,4,4)$$

$n=5$: $$(2, 4, 8, 16, 16), (2, 4, 12, 12, 12), (2, 6, 6, 12, 12), (2, 6, 8, 10, 16), (2, 6, 10, 10, 10), (2, 8, 8, 8, 8), (3, 3, 6, 12, 12), (3, 3, 9, 9, 9), (3, 4, 5, 14, 14), (3, 4, 6, 10, 16), (3, 4, 6, 11, 11), (3, 4, 8, 8, 8), (3, 5, 5, 9, 9), (3, 5, 6, 7, 12), (3, 5, 7, 7, 9), (3, 5, 7, 8, 8), (3, 6, 6, 6, 6), (4, 4, 4, 8, 8), (4, 4, 5, 7, 12), (4, 4, 6,6, 6), (4, 5, 5, 6, 10), (4, 5, 5, 7, 7), (5, 5, 5, 5, 5)$$

$n=6$: $$(2, 4, 8, 16, 32, 32), (2, 4, 8, 24, 24, 24), (2, 4, 12, 12, 24, 24), (2, 4, 12, 20, 20, 20), (2, 4, 16, 16, 16, 16), (2, 6, 6, 12, 24, 24), (2, 6, 6, 18, 18, 18), (2, 6, 8, 10, 28, 28), (2, 6, 8, 12, 20, 32), (2, 6, 8, 12, 22, 22), (2, 6, 8, 16, 16, 16), (2, 6, 12, 12, 12, 12), (2, 10, 10, 10, 10, 10), (3, 3, 6, 12, 24, 24), (3, 3, 6, 18, 18, 18), (3, 3, 9, 9, 18, 18), (3, 3, 9, 15, 15, 15), (3, 3, 12, 12, 12, 12), (3, 4, 5, 14, 28, 28), (3, 4, 5, 22, 22, 22), (3, 4, 6, 10, 32, 32), (3, 4, 6, 11, 21, 32), (3, 4, 6, 11, 22, 22), (3, 4, 6, 16, 16, 16), (3, 4, 8, 8, 16, 16), (3, 4, 8, 13, 13, 13), (3, 4, 11, 11, 11, 11), (3, 5, 5, 9, 18, 18), (3, 5, 5, 14, 14, 14), (3, 5, 6, 7, 24, 24), (3, 5, 6, 8, 21, 21), (3, 5, 6, 9, 15, 24), (3, 5, 6, 9, 17, 17), (3, 5, 6, 12, 12, 12), (3, 5, 9, 9, 9, 9), (3, 8, 8, 8, 8, 8), (4, 4, 4, 8, 16, 16), (4, 4, 4, 12, 12, 12), (4, 4, 6, 6, 12, 12), (4, 4, 6, 10, 10, 10), (4, 4, 8, 8, 8, 8), (4, 7, 7, 7, 7, 7), (6, 6, 6, 6, 6, 6)$$

I can't see any clear pattern which would allow a simpler solution, but it's definitely possible one exists.

For a particular tuple of values, we test it by constructing a directed graph of all tuples smaller then it, element-wise, representing the distance from the last element of each series, with the edges being "increase all values by 1, except for one which is set to 0", and then testing whether there's a cycle in the graph.

For example, for $(2, 4, 4)$, we get this graph:

Note the $(1,0,2), (0, 1, 3), (1, 2,0), (0, 3, 1)$ cycle, which gives the solution $1312$ repeating (each number is the index of the series the value occurs in)

This immediately proves that $\sum_i {\frac1{N_i}} \leq 1$ is necessary, otherwise $f(p) = \sum_i {\frac{p_i}{N_i}}$ gives a topological sort of the graph.

I have programmed this, and for $n \leq 5$, a tuple $N_1, N_2, ..., N_n$ is valid if it's elementwise greater than or equal to any of the following:

$n=1$: $$(2)$$

$n=2$: $$(2, 2)$$

$n=3$: $$(2, 4, 4), (3, 3, 3)$$

$n=4$: $$(2, 4, 8, 8), (2,6,6,6),(3,3,6,6),(3,4,5,8),(3,5,5,5),(4,4,4,4)$$

$n=5$: $$(2, 4, 8, 16, 16), (2, 4, 12, 12, 12), (2, 6, 6, 12, 12), (2, 6, 8, 10, 16), (2, 6, 10, 10, 10), (2, 8, 8, 8, 8), (3, 3, 6, 12, 12), (3, 3, 9, 9, 9), (3, 4, 5, 14, 14), (3, 4, 6, 10, 16), (3, 4, 6, 11, 11), (3, 4, 8, 8, 8), (3, 5, 5, 9, 9), (3, 5, 6, 7, 12), (3, 5, 7, 7, 9), (3, 5, 7, 8, 8), (3, 6, 6, 6, 6), (4, 4, 4, 8, 8), (4, 4, 5, 7, 12), (4, 4, 6,6, 6), (4, 5, 5, 6, 10), (4, 5, 5, 7, 7), (5, 5, 5, 5, 5)$$

I can't see any clear pattern. I am currently running the algorithm for $n=6$, but since it takes exponential time it will take a long time.

For a particular tuple of values, we test it by constructing a directed graph of all tuples smaller then it, element-wise, representing the distance from the last element of each series, with the edges being "increase all values by 1, except for one which is set to 0", and then testing whether there's a cycle in the graph.

For example, for $(2, 4, 4)$, we get this graph:

Note the $(1,0,2), (0, 1, 3), (1, 2,0), (0, 3, 1)$ cycle, which gives the solution $1312$ repeating (each number is the index of the series the value occurs in)

This immediately proves that $\sum_i {\frac1{N_i}} \leq 1$ is necessary, otherwise $f(p) = \sum_i {\frac{p_i}{N_i}}$ gives a topological sort of the graph.

I have programmed this, and for $n \leq 5$, a tuple $N_1, N_2, ..., N_n$ is valid if it's elementwise greater than or equal to any of the following:

$n=1$: $$(2)$$

$n=2$: $$(2, 2)$$

$n=3$: $$(2, 4, 4), (3, 3, 3)$$

$n=4$: $$(2, 4, 8, 8), (2,6,6,6),(3,3,6,6),(3,4,5,8),(3,5,5,5),(4,4,4,4)$$

$n=5$: $$(2, 4, 8, 16, 16), (2, 4, 12, 12, 12), (2, 6, 6, 12, 12), (2, 6, 8, 10, 16), (2, 6, 10, 10, 10), (2, 8, 8, 8, 8), (3, 3, 6, 12, 12), (3, 3, 9, 9, 9), (3, 4, 5, 14, 14), (3, 4, 6, 10, 16), (3, 4, 6, 11, 11), (3, 4, 8, 8, 8), (3, 5, 5, 9, 9), (3, 5, 6, 7, 12), (3, 5, 7, 7, 9), (3, 5, 7, 8, 8), (3, 6, 6, 6, 6), (4, 4, 4, 8, 8), (4, 4, 5, 7, 12), (4, 4, 6,6, 6), (4, 5, 5, 6, 10), (4, 5, 5, 7, 7), (5, 5, 5, 5, 5)$$

$n=6$: $$(2, 4, 8, 16, 32, 32), (2, 4, 8, 24, 24, 24), (2, 4, 12, 12, 24, 24), (2, 4, 12, 20, 20, 20), (2, 4, 16, 16, 16, 16), (2, 6, 6, 12, 24, 24), (2, 6, 6, 18, 18, 18), (2, 6, 8, 10, 28, 28), (2, 6, 8, 12, 20, 32), (2, 6, 8, 12, 22, 22), (2, 6, 8, 16, 16, 16), (2, 6, 12, 12, 12, 12), (2, 10, 10, 10, 10, 10), (3, 3, 6, 12, 24, 24), (3, 3, 6, 18, 18, 18), (3, 3, 9, 9, 18, 18), (3, 3, 9, 15, 15, 15), (3, 3, 12, 12, 12, 12), (3, 4, 5, 14, 28, 28), (3, 4, 5, 22, 22, 22), (3, 4, 6, 10, 32, 32), (3, 4, 6, 11, 21, 32), (3, 4, 6, 11, 22, 22), (3, 4, 6, 16, 16, 16), (3, 4, 8, 8, 16, 16), (3, 4, 8, 13, 13, 13), (3, 4, 11, 11, 11, 11), (3, 5, 5, 9, 18, 18), (3, 5, 5, 14, 14, 14), (3, 5, 6, 7, 24, 24), (3, 5, 6, 8, 21, 21), (3, 5, 6, 9, 15, 24), (3, 5, 6, 9, 17, 17), (3, 5, 6, 12, 12, 12), (3, 5, 9, 9, 9, 9), (3, 8, 8, 8, 8, 8), (4, 4, 4, 8, 16, 16), (4, 4, 4, 12, 12, 12), (4, 4, 6, 6, 12, 12), (4, 4, 6, 10, 10, 10), (4, 4, 8, 8, 8, 8), (4, 7, 7, 7, 7, 7), (6, 6, 6, 6, 6, 6)$$

I can't see any clear pattern which would allow a simpler solution, but it's definitely possible one exists.

For a particular tuple of values, we test it by constructing a directed graph of all tuples smaller then it, element-wise, representing the distance from the last element of each series, with the edges being "increase all values by 1, except for one which is set to 0", and then testing whether there's a cycle in the graph.

This immediately proves that $\sum_i {\frac1{N_i}} \leq 1$ is necessary, otherwise $f(p) = \sum_i {\frac{p_i}{N_i}}$ gives a topological sort of the graph.

I have programmed this, and for $n \leq 5$, a tuple $N_1, N_2, ..., N_n$ is valid if it's elementwise greater than or equal to any of the following:

$n=1$: $$(2)$$

$n=2$: $$(2, 2)$$

$n=3$: $$(2, 4, 4), (3, 3, 3)$$

$n=4$: $$(2, 4, 8, 8), (2,6,6,6),(3,3,6,6),(3,4,5,8),(3,5,5,5),(4,4,4,4)$$

$n=5$: $$(2, 4, 8, 16, 16), (2, 4, 12, 12, 12), (2, 6, 6, 12, 12), (2, 6, 8, 10, 16), (2, 6, 10, 10, 10), (2, 8, 8, 8, 8), (3, 3, 6, 12, 12), (3, 3, 9, 9, 9), (3, 4, 5, 14, 14), (3, 4, 6, 10, 16), (3, 4, 6, 11, 11), (3, 4, 8, 8, 8), (3, 5, 5, 9, 9), (3, 5, 6, 7, 12), (3, 5, 7, 7, 9), (3, 5, 7, 8, 8), (3, 6, 6, 6, 6), (4, 4, 4, 8, 8), (4, 4, 5, 7, 12), (4, 4, 6,6, 6), (4, 5, 5, 6, 10), (4, 5, 5, 7, 7), (5, 5, 5, 5, 5)$$

I can't see any clear pattern. I am currently running the algorithm for $n=6$, but since it takes exponential time it will take a long time.

For a particular tuple of values, we test it by constructing a directed graph of all tuples smaller then it, element-wise, representing the distance from the last element of each series, with the edges being "increase all values by 1, except for one which is set to 0", and then testing whether there's a cycle in the graph.

For example, for $(2, 4, 4)$, we get this graph:

Note the $(1,0,2), (0, 1, 3), (1, 2,0), (0, 3, 1)$ cycle, which gives the solution $1312$ repeating (each number is the index of the series the value occurs in)

This immediately proves that $\sum_i {\frac1{N_i}} \leq 1$ is necessary, otherwise $f(p) = \sum_i {\frac{p_i}{N_i}}$ gives a topological sort of the graph.

I have programmed this, and for $n \leq 5$, a tuple $N_1, N_2, ..., N_n$ is valid if it's elementwise greater than or equal to any of the following:

$n=1$: $$(2)$$

$n=2$: $$(2, 2)$$

$n=3$: $$(2, 4, 4), (3, 3, 3)$$

$n=4$: $$(2, 4, 8, 8), (2,6,6,6),(3,3,6,6),(3,4,5,8),(3,5,5,5),(4,4,4,4)$$

$n=5$: $$(2, 4, 8, 16, 16), (2, 4, 12, 12, 12), (2, 6, 6, 12, 12), (2, 6, 8, 10, 16), (2, 6, 10, 10, 10), (2, 8, 8, 8, 8), (3, 3, 6, 12, 12), (3, 3, 9, 9, 9), (3, 4, 5, 14, 14), (3, 4, 6, 10, 16), (3, 4, 6, 11, 11), (3, 4, 8, 8, 8), (3, 5, 5, 9, 9), (3, 5, 6, 7, 12), (3, 5, 7, 7, 9), (3, 5, 7, 8, 8), (3, 6, 6, 6, 6), (4, 4, 4, 8, 8), (4, 4, 5, 7, 12), (4, 4, 6,6, 6), (4, 5, 5, 6, 10), (4, 5, 5, 7, 7), (5, 5, 5, 5, 5)$$

I can't see any clear pattern. I am currently running the algorithm for $n=6$, but since it takes exponential time it will take a long time.